A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?A. `(1)/(3)`B. 2C. 3D. 1

A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in

1.	A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?A. `(1)/(3)`B. 2C. 3D. 1
Answer» Correct Answer - A::D `(dy)/(dx)=(2t)/(1+3t^(2))=(1)/(2)` `rArr" "3t^(2)-4t+1=0`

The derivative of `cos(2tan^(-1)sqrt((1-x)/(1+x)))-2cos^(-1)sqrt((1-x)/(2))` w.r.t. x isA. `1-(1)/(sqrt(1-x^(2)))`B. `1-(1)/(sqrt(1+x^(2)))`C. `2-(1)/(sqrt(1-x^(2)))`D. `2-(1)/(sqrt(1+x^(2)))`
`(d)/(dx)[cos^(-1)(xsqrtx-sqrt((1-x)(1-x^(2))))]=`A. `(1)/(sqrt(1-x^(2)))-(1)/(2sqrt(x-x^(2)))`B. `(-1)/(sqrt(1-x^(2)))-(1)/(2sqrt(x-x^(2)))`C. `(1)/(sqrt(1-x^(2)))+(1)/(2sqrt(x-x^(2)))`D. `(1)/(sqrt(1-x^(2)))`
Let `y=x^3-8x+7a n dx=f(t)dot`If `(dy)/(dt)=2`and `x=3`at `t=0,`then `(dx)/(dt)`at `t=0`is given by1 (b)`(19)/2`(c) `2/(19)`(d) none of theseA. 1B. `(19)/(2)`C. `(2)/(19)`D. None of these
if `x=(1+t)/t^3 ,y=3/(2t^2)+2/t` satisfies `f(x)*{(dy)/(dx)}^3=1+(dy)/(dx)` then `f(x)` is:A. `x`B. `(x^(2))/(1+X^(2))`C. `x+x+(1)/(x)`D. `x-(1)/(x)`
If `x=sectheta-costheta` and `y=sec^n theta- cos^n theta` then show that `(x^2+4)((dy)/(dx))^2=n^2(y^2+4)`A. 8B. 16C. 64D. 49
A curve parametrically given by `x=t+t^(3)" and "y=t^(2)," where "t in R." For what vlaue(s) of t is "(dy)/(dx)=(1)/(2)`?A. `(1)/(3)`B. 2C. 3D. 1
If `y^3-y=2x ,t h e n(x^2-1/(27))(d^2y)/(dx^2)+x(dy)/d=``y`b. `y/3`c. `y/9`d. `y/(27)`A. yB. `(y)/(3)`C. `(y)/(9)`D. `(y)/(27)`